1

I have two limit problems which are quite similar, so I've put them both into this post.

  1. $((4^{10}+2^{n})^{\frac{1}{n}})$
  2. $((3n^{2}+n)^{\frac{1}{n}})$

Attempt:

For 1, I'm not sure if this is permissible. I know that $\lim_{n \to \infty}(x^{n}+y^{n})^{\frac{1}{n}} = \max\{x,y\}$. Obviously $4^{10}$ is a constant so it's not quite in the same form, but I believe it is still okay to use this result, and so the limit is $2$.

For 2, I've got $((3n^{2}+n)^{\frac{1}{n}}) = (n(3n+1)^{\frac{1}{n}})$, but then I'm not too sure how to proceed.

I also thought of taking the log and proceeding that way, however in the notes I'm learning from they haven't covered l'hopital yet, so I'm trying to not use it.

Thanks.

AnalysisLearner
  • 319

2 Answers2

2

For the first one we have

$$(4^{10}+2^{n})^{\frac{1}{n}}=2\left(\frac{4^{10}}{2^n}+1\right)^{\frac{1}{n}}\to 2\cdot 1^0 =2$$

and

$$(3n^{2}+n)^{\frac{1}{n}}=e^{\log(3n^{2}+n)^{\frac{1}{n}}}=e^{\frac{\log (3n^2+n) }{n}}\to e^0= 1$$

recall indeed that $\frac{\log n}n \to 0$ and therefore

$$\frac{\log (3n^2+n) }{n}\le \frac{\log (3n^2+6n+3) }{n}=\frac{\log (3(n+1)^2) }{n}=\frac{2\log (n+1)+\log 3 }{n}=$$

$$=2\frac{n+1}n\frac{\log (n+1) }{n+1}+\frac{\log 3}n\to 2 \cdot 1 \cdot 0 + 0 =0$$

As an alternative for the second we can also use that

$$n^\frac1n \le(3n^{2}+n)^{\frac{1}{n}} \le (3n^{2}+n^2)^{\frac{1}{n}}=(4n^{2})^{\frac{1}{n}}$$

and conclude in a similar way by squeeze theorem.

user
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  • For the first one, where do you get the factor of $2$ from in the first equality? For the second one where does the $2$ in $2 \frac{n+1}{n}$ come from? As an aside, how would one prove the result about $ \frac{\log n}{n} \to 0$ (I'll have a go myself using definition of null sequence / convergence). Thanks (again!). – AnalysisLearner Nov 04 '19 at 16:14
  • @AnalysisLearner It is a simple step $$(4^{10}+2^{n})^{\frac{1}{n}}=\left(2^n \left(\frac{4^{10}}{2^n}+1\right)\right)^{\frac{1}{n}}=(2^n)^\frac1n\left(\frac{4^{10}}{2^n}+1\right)^{\frac{1}{n}}=2\left(\frac{4^{10}}{2^n}+1\right)^{\frac{1}{n}}$$ – user Nov 04 '19 at 16:53
  • @AnalysisLearner ALso the second one is a simple step $$\frac{2\log (n+1)}{n}=2\frac{n+1}n\frac{\log (n+1) }{n+1}$$ I've noticed now tere is a small typo, I fix that. – user Nov 04 '19 at 16:55
  • @AnalysisLearner A way to prove that is by $ \frac{\log x}{x} \to 0$ which by $x=e^y \to \infty$ becomes $ \frac{\log e^y}{e^y}=\frac y{e^y} \to 0$ . – user Nov 04 '19 at 16:59
  • @AnalysisLearner Refer also to the related OP. – user Nov 04 '19 at 17:01
  • that makes sense now. Thanks. – AnalysisLearner Nov 04 '19 at 17:01
1

Take $\varepsilon>0$. Then $\lim_{n\to\infty}(2+\varepsilon)^n=\infty$ and therefroe $(2+\varepsilon)^n>4^{10}$, if $n\gg1$. But then, if $n$ is large enough,$$(2^n)^{1/n}<(4^{10}+2^n)^{1/n}<\bigl((2+\varepsilon)^n+2^n\bigr)^{1/n}.$$Since $\lim_{n\to\infty}(2^n)^{1/n}=2$ and$$\lim_{n\to\infty}\bigl((2+\varepsilon)^n+2^n\bigr)^{1/n}=\max\{2+\varepsilon,2\}=2+\varepsilon,$$it is easy now to prove that the limit of your sequence is $2$.

For the other sequence, you can use the fact that $3n^2<3n^2+n<4n^2$ and that$$\lim_{n\to\infty}(3n^2)^{1/n}=\lim_{n\to\infty}(4n^2)^{1/n}=1.$$

José Carlos Santos
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